Question: Simplify and expand the following expression: $ \dfrac{5}{5q - 40}+ \dfrac{5}{3q - 12}+ \dfrac{3}{q^2 - 12q + 32} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{5}{5q - 40} = \dfrac{5}{5(q - 8)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3q - 12} = \dfrac{5}{3(q - 4)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 - 12q + 32} = \dfrac{3}{(q - 8)(q - 4)}$ Now we have: $ \dfrac{5}{5(q - 8)}+ \dfrac{5}{3(q - 4)}+ \dfrac{3}{(q - 8)(q - 4)} $ The least common multiple of the denominators is: $ 15(q - 8)(q - 4)$ In order to get the first term over $15(q - 8)(q - 4)$ , multiply by $\dfrac{3(q - 4)}{3(q - 4)}$ $ \dfrac{5}{5(q - 8)} \times \dfrac{3(q - 4)}{3(q - 4)} = \dfrac{15(q - 4)}{15(q - 8)(q - 4)} $ In order to get the second term over $15(q - 8)(q - 4)$ , multiply by $\dfrac{5(q - 8)}{5(q - 8)}$ $ \dfrac{5}{3(q - 4)} \times \dfrac{5(q - 8)}{5(q - 8)} = \dfrac{25(q - 8)}{15(q - 8)(q - 4)} $ In order to get the third term over $15(q - 8)(q - 4)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{3}{(q - 8)(q - 4)} \times \dfrac{15}{15} = \dfrac{45}{15(q - 8)(q - 4)} $ Now we have: $ \dfrac{15(q - 4)}{15(q - 8)(q - 4)} + \dfrac{25(q - 8)}{15(q - 8)(q - 4)} + \dfrac{45}{15(q - 8)(q - 4)} $ $ = \dfrac{ 15(q - 4) + 25(q - 8) + 45} {15(q - 8)(q - 4)} $ Expand: $ = \dfrac{15q - 60 + 25q - 200 + 45}{15q^2 - 180q + 480} $ $ = \dfrac{40q - 215}{15q^2 - 180q + 480}$ Simplify: $ = \dfrac{8q - 43}{3q^2 - 36q + 96}$